Problem: Solve for $y$, $ \dfrac{8}{y - 1} = \dfrac{6}{4y - 4} + \dfrac{4y}{y - 1} $
First we need to find a common denominator for all the expressions. This means finding the least common multiple of $y - 1$ $4y - 4$ and $y - 1$ The common denominator is $4y - 4$ To get $4y - 4$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ \dfrac{8}{y - 1} \times \dfrac{4}{4} = \dfrac{32}{4y - 4} $ The denominator of the second term is already $4y - 4$ , so we don't need to change it. To get $4y - 4$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{4y}{y - 1} \times \dfrac{4}{4} = \dfrac{16y}{4y - 4} $ This give us: $ \dfrac{32}{4y - 4} = \dfrac{6}{4y - 4} + \dfrac{16y}{4y - 4} $ If we multiply both sides of the equation by $4y - 4$ , we get: $ 32 = 6 + 16y$ $ 32 = 16y + 6$ $ 26 = 16y $ $ y = \dfrac{13}{8}$